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Consider the following question:

Let $f \in \operatorname{RS}_a^b (\alpha)$, that is, the function $f$ is Riemann-Stieltjes integrable over the interval $[a,b]$ with respect to $\alpha = \alpha(x)$, and let $\alpha$ be continuous. Prove that $$\left\{ \int_a^{b-\frac{1}{n}} f(x) d\alpha(x) \right\}_{n\in\mathbb{N}} \to \int_a^{b} f(x) d\alpha(x)$$

What I tried to do was to consider an arbitrary $\epsilon > 0$ and show that $$\left| \int_a^{b-\frac{1}{n}} f d\alpha-\int_a^{b} f d\alpha\right|=\left| - \int_{b-\frac{1}{n}}^b f d\alpha\right|=\left| \int_{b-\frac{1}{n}}^b f d\alpha\right|< \epsilon$$ Now, intuitively this last term tends to zero as $n \to \infty$, so we could always pick a big enough $n$ to make it smaller than $\epsilon$. How can I argue this formally? Is this the correct way to proceed? I feel like I am missing something.

Bergson
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    Hint: $f$ is bounded and $\alpha$ is (uniformly) continuous. – Matematleta May 19 '19 at 19:19
  • @Matematleta how do we know that $f$ is bounded? Wouldn't we need it to be continuous to say that? – Bergson May 19 '19 at 19:29
  • It's part of the definition of the R-S integral that $f$ be bounded, right? – Matematleta May 19 '19 at 19:31
  • Boundedness is not just part of the definition, it is a necessary condition for RS integrability. – RRL May 19 '19 at 19:54
  • @Matematleta So, assuming we can say that $f$ is bounded (say by $K$) then $$\left| \int_{b-\frac{1}{n}}^b f d\alpha\right| \leq \left| K \int_{b-\frac{1}{n}}^b d\alpha\right| $$ and then writing it as a sum over a partition we get $$|K \sum_P \Delta \alpha_i| = |K(\alpha(b)-\alpha(b-\frac{1}{n}))|$$ which can be made as small as we want by the uniform continuity of $\alpha$. Does this look right to you? – Bergson May 19 '19 at 19:56
  • @ThomasBladt: That's fine as long as $\alpha$ is monotone. – RRL May 19 '19 at 19:57
  • @RRL indeed, and that is what I meant. – Matematleta May 19 '19 at 20:01
  • Ok, thank you to both :) – Bergson May 19 '19 at 20:03
  • It is not yet clear how to prove this if $\alpha$ has unbounded variation. There are such functions that are also continuous on compact intervals. But it is rare for the RS integral to exist unless for example $f$ is constant. – RRL May 19 '19 at 20:05

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